Bird J. Electrical Circuit Theory and Technology

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346 Electrical Circuit Theory and Technology

Equivalent circuit of a transformer

18 A transformer has 1200 primary turns and 200 secondary turns. The

primary and secondary resistances are 0.2  and 0.02  respectively

and the corresponding leakage reactances are 1.2  and 0.05 

respectively. Calculate (a) the equivalent resistance, reactance and

impedance referred to the primary winding, and (b) the phase angle

of the impedance. [(a) 0.92 ,3.0,3.14 (b) 72.95

]

Regulation

19 A 6 kVA, 100 V/500 V, single-phase transformer has a secondary

terminal voltage of 487.5 volts when loaded. Determine the regula-

tion of the transformer. [2.5%]

20 A transformer has an open circuit voltage of 110 volts. A tap-

changing device operates when the regulation falls below 3%.

Calculate the load voltage at which the tap-changer operates.

[106.7 volts]

Losses and efﬁciency

21 A single-phase transformer has a voltage ratio of 6:1 and the h.v.

winding is supplied at 540 V. The secondary winding provides a full

load current of 30 A at a power factor of 0.8 lagging. Neglecting

losses, ﬁnd (a) the rating of the transformer, (b) the power supplied

to the load, (c) the primary current.

[(a) 2.7 kVA, (b) 2.16 kW, (c) 5 A]

22 A single-phase transformer is rated at 40 kVA. The transformer has

full-load copper losses of 800 W and iron losses of 500 W. Deter-

mine the transformer efﬁciency at full load and 0.8 power factor.

[96.10%]

23 Determine the efﬁciency of the transformer in problem 22 at half

full-load and 0.8 power factor. [95.81%]

24 A 100 kVA, 2000 V/400 V, 50 Hz, single-phase transformer has an

iron loss of 600 W and a full-load copper loss of 1600 W. Calculate

its efﬁciency for a load of 60 kW at 0.8 power factor. [97.56%]

25 (a) What are eddy currents? State how their effect is reduced in

transformers.

(b) Determine the efﬁciency of a 15 kVA transformer for the

following conditions:

(i) full-load, unity power factor

(ii) 0.8 full-load, unity power factor

(iii) half full-load, 0.8 power factor.

Transformers 347

Assume that iron losses are 200 W and the full-load copper loss is

300 W. [(a) 96.77% (ii) 96.84% (iii) 95.62%]

26 A 250 kVA transformer has a full load copper loss of 3 kW and

an iron loss of 2 kW. Calculate (a) the output kVA at which the

efﬁciency of the transformer is a maximum, and (b) the maximum

efﬁciency, assuming the power factor of the load is 0.80.

[(a) 204.1 kVA (b) 97.61%]

Resistance matching

27 A transformer having a turns ratio of 8:1 supplies a load of resistance

50 . Determine the equivalent input resistance of the transformer.

[3.2 k]

28 What ratio of transformer is required to make a load of resistance

30  appear to have a resistance of 270 ? [3:1]

29 A single-phase, 240 V/2880 V ideal transformer is supplied from a

240 V source through a cable of resistance 3 . If the load across

the secondary winding is 720  determine (a) the primary current

ﬂowing and (b) the power dissipated in the load resistance.

[(a) 30 A (b) 4.5 kW]

30 A load of resistance 768  is to be matched to an ampliﬁer which

has an effective output resistance of 12 . Determine the turns ratio

of the coupling transformer. [1:8]

31 An a.c. source of 20 V and internal resistance 20 k is matched to

a load by a 16:1 single-phase transformer. Determine (a) the value

of the load resistance and (b) the power dissipated in the load.

[(a) 78.13  (b) 5 mW]

Auto-transformer

32 A single-phase auto transformer has a voltage ratio of 480 V:300 V

and supplies a load of 30 kVA at 300 V. Assuming an ideal trans-

former, calculate the current in each section of the winding.

D 62.5 A, I

D 100 A, I

 I

 D 37.5 A]

33 Calculate the saving in the volume of copper used in an auto

transformer compared with a double-wound transformer for (a) a

300 V:240 V transformer, and (b) a 400 V:100 V transformer.

[(a) 80% (b) 25%]

Three-phase transformer

34 A three-phase transformer has 600 primary turns and 150 secondary

turns. If the supply voltage is 1.5 kV determine the secondary line

voltage on no-load when the windings are connected (a) delta-star,

(b) star-delta. [(a) 649.5 V (b) 216.5 V]

348 Electrical Circuit Theory and Technology

Current transformer

35 A current transformer has two turns on the primary winding and a

secondary winding of 260 turns. The secondary winding is connected

to an ammeter with a resistance of 0.2 . The resistance of the

secondary winding is 0.3 . If the current in the primary winding is

650 A, determine (a) the reading on the ammeter, (b) the potential

difference across the ammeter, and (c) the total load in VA on the

secondary. [(a

)5A(b)1V(c)7.5VA]

Assignment 6

This assignment covers the material contained in chapters 19

and 20.

The marks for each question are shown in brackets at the end of

each question.

1 Three identical coils each of resistance 40  and inductive reactance

30  are connected (i) in star, and (ii) in delta to a 400 V, three-phase

supply. Calculate for each connection (a) the line and phase voltages,

(b) the phase and line currents, and (c) the total power dissipated.

(12)

2 Two wattmeters are connected to measure the input power to a

balanced three-phase load by the two-wattmeter method. If the

instrument readings are 10 kW and 6 kW, determine (a) the total

power input, and (b) the load power factor. (5)

3 An ideal transformer connected to a 250 V mains, supplies a 25 V,

200 W lamp. Calculate the transformer turns ratio and the current

taken from the supply. (5)

4 A 200 kVA, 8000 V/320 V, 50 Hz single phase transformer has 120

secondary turns. Determine (a) the primary and secondary currents,

(b) the number of primary turns, and (c) the maximum value of ﬂux.

(9)

5 Determine the percentage regulation of an 8 kVA, 100 V/200 V,

single phase transformer when it’s secondary terminal voltage is

194 V when loaded. (3)

6 A 500 kVA rated transformer has a full-load copper loss of 4 kW and

an iron loss of 3 kW. Determine the transformer efﬁciency (a) at full

load and 0.80 power factor, and (b) at half full load and 0.80 power

factor. (10)

7 Determine the optimum value of load resistance for maximum power

transfer if the load is connected to an ampliﬁer of output resistance

288  through a transformer with a turns ratio 6:1. (3)

8 A single-phase auto transformer has a voltage ratio of 250 V:200 V

and supplies a load of 15 kVA at 200 V. Assuming an ideal trans-

former, determine the current in each section of the winding. (3)

21 D.c. machines

At the end of this chapter you should be able to:

ž distinguish between the function of a motor and a generator

ž describe the action of a commutator

ž describe the construction of a d.c. machine

ž distinguish between wave and lap windings

ž understand shunt, series and compound windings of d.c.

machines

ž understand armature reaction

ž calculate generated e.m.f. in an armature winding using

E D

2pnZ

ž describe types of d.c. generator and their characteristics

ž calculate generated e.m.f. for a generator using E D V C I

ž state typical applications of d.c. generators

ž list d.c. machine losses and calculate efﬁciency

ž calculate back e.m.f. for a d.c. motor using E D V  I

ž calculate the torque of a d.c. motor using T D

2n

and

T D

pZI

c

ž describe types of d.c. motor and their characteristics

ž state typical applications of d.c. motors

ž describe a d.c. motor starter

ž describe methods of speed control of d.c. motors

ž list types of enclosure for d.c. motors

21.1 Introduction

When the input to an electrical machine is electrical energy, (seen as

applying a voltage to the electrical terminals of the machine), and the

output is mechanical energy, (seen as a rotating shaft), the machine is

called an electric motor. Thus an electric motor converts electrical energy

into mechanical energy.

The principle of operation of a motor is explained in Section 8.4,

page 96.

When the input to an electrical machine is mechanical energy, (seen as,

say, a diesel motor, coupled to the machine by a shaft), and the output is

D.c. machines 351

electrical energy, (seen as a voltage appearing at the electrical terminals

of the machine), the machine is called a generator. Thus, a generator

converts mechanical energy to electrical energy.

The principle of operation of a generator is explained in

Section 9.2, page 101.

21.2 The action of a

commutator

In an electric motor, conductors rotate in a uniform magnetic ﬁeld. A

single-loop conductor mounted between permanent magnets is shown in

Figure 21.1. A voltage is applied at points A and B in Figure 21.1(a).

Figure 21.1

A force, F, acts on the loop due to the interaction of the magnetic

ﬁeld of the permanent magnets and the magnetic ﬁeld created by the

current ﬂowing in the loop. This force is proportional to the ﬂux density,

B, the current ﬂowing, I, and the effective length of the conductor, l,

i.e. F D BIl. The force is made up of two parts, one acting vertically

downwards due to the current ﬂowing from C to D and the other acting

vertically upwards due to the current ﬂowing from E to F (from Fleming’s

left hand rule). If the loop is free to rotate, then when it has rotated

through 180

, the conductors are as shown in Figure 21.1(b). For rotation

to continue in the same direction, it is necessary for the current ﬂow

to be as shown in Figure 21.1(b), i.e. from D to C and from F to E.

This apparent reversal in the direction of current ﬂow is achieved by a

process called commutation. With reference to Figure 21.2(a), when a

direct voltage is applied at A and B, then as the single-loop conductor

rotates, current ﬂow will always be away from the commutator for the

part of the conductor adjacent to the N-pole and towards the commutator

for the part of the conductor adjacent to the S-pole. Thus the forces act to

give continuous rotation in an anti-clockwise direction. The arrangement

shown in Figure 21.2(a) is called a ‘two-segment’ commutator and the

voltage is applied to the rotating segments by stationary brushes, (usually

carbon blocks), which slide on the commutator material, (usually copper),

when rotation takes place.

In practice, there are many conductors on the rotating part of a

d.c. machine and these are attached to many commutator segments.

A schematic diagram of a multi-segment commutator is shown in

Figure 21.2(b).

352 Electrical Circuit Theory and Technology

Figure 21.2

Poor commutation results in sparking at the trailing edge of the brushes.

This can be improved by using interpoles (situated between each pair of

main poles), high resistance brushes, or using brushes spanning several

commutator segments.

21.3 D.c. machine

construction

The basic parts of any d.c. machine are shown in Figure 21.3, and

comprise:

(a) a stationary part called the stator having,

(i) a steel ring called the yoke, to which are attached

(ii) the magnetic poles, around which are the

(iii) ﬁeld windings, i.e. many turns of a conductor wound round

the pole core; current passing through this conductor creates

an electromagnet, (rather than the permanent magnets shown

in Figures 21.1 and 21.2),

(b) a rotating part called the armature mounted in bearings housed in

the stator and having,

Figure 21.3

(iv) a laminated cylinder of iron or steel called the core,onwhich

teeth are cut to house the

(v) armature winding, i.e. a single or multi-loop conductor

system and

(vi) the commutator, (see Section 21.2).

Armature windings can be divided into two groups, depending on how

the wires are joined to the commutator. These are called wave windings

and lap windings.

(a) In wave windings there are two paths in parallel irrespective of the

number of poles, each path supplying half the total current output.

Wave wound generators produce high voltage, low current outputs.

(b) In lap windings there are as many paths in parallel as the machine

has poles. The total current output divides equally between them.

Lap wound generators produce high current, low voltage output.

D.c. machines 353

21.4 Shunt, series and

compound windings

When the ﬁeld winding of a d.c. machine is connected in parallel with

the armature, as shown in Figure 21.4(a), the machine is said to be shunt

wound. If the ﬁeld winding is connected in series with the armature, as

shown in Figure 21.4(b), then the machine is said to be series wound.

A compound wound machine has a combination of series and shunt

windings.

Depending on whether the electrical machine is series wound, shunt

wound or compound wound, it behaves differently when a load is applied.

The behaviour of a d.c. machine under various conditions is shown by

means of graphs, called characteristic curves or just characteristics. The

characteristics shown in the following sections are theoretical, since they

neglect the effects of armature reaction.

Armature reaction is the effect that the magnetic ﬁeld produced by the

armature current has on the magnetic ﬁeld produced by the ﬁeld system.

In a generator, armature reaction results in a reduced output voltage, and

in a motor, armature reaction results in increased speed.

A way of overcoming the effect of armature reaction is to ﬁt compen-

sating windings, located in slots in the pole face.

Figure 21.4

21.5 E.m.f. generated in

an armature winding

Let Z D number of armature conductors,

 D useful ﬂux per pole, in webers

p D number of pairs of poles

and n D armature speed in rev/s

The e.m.f. generated by the armature is equal to the e.m.f. generated by

one of the parallel paths. Each conductor passes 2p poles per revolution

and thus cuts 2p webers of magnetic ﬂux per revolution. Hence ﬂux cut

by one conductor per second D 2pn Wb and so the average e.m.f. E

generated per conductor is given by:

E D 2pn volts since 1 volt D 1 Weber per second

Let c D number of parallel paths through the winding between positive

and negative brushes

= 2 for a wave winding

= 2p for a lap winding

The number of conductors in series in each path D

The total e.m.f. between brushes

D (average e.m.f./conductor)(number of conductors in series per path)

D 2pn

354 Electrical Circuit Theory and Technology

i.e.,

generated e.m.f., E =

2p8nZ

volts

21.1

Since Z, p and c are constant for a given machine, then E / n.

However 2n is the angular velocity ω in radians per second, hence the

generated e.m.f. is proportional to  and ω, i.e.,

generated e.m.f., E ∝ 8!

21.2

Problem 1. An 8-pole, wave-connected armature has 600 conduc-

tors and is driven at 625 rev/min. If the ﬂux per pole is 20 mWb,

determine the generated e.m.f.

Z D 600, c D 2 (for a wave winding), p D 4 pairs

n D

625

rev/s,  D 20 ð 10

3

Generated e.m.f., E D

2pnZ

2420 ð 10

3





625



600

D 500 volts

Problem 2. A 4-pole generator has a lap-wound armature with

50 slots with 16 conductors per slot. The useful ﬂux per pole is

30 mWb. Determine the speed at which the machine must be driven

to generate an e.m.f. of 240 V.

E D 240 V, c D 2p (for a lap winding), Z D 50 ð 16 D 800,

 D 30 ð 10

3

Wb.

Generated e.m.f. E D

2pnZ

D nZ

Rearranging gives, speed, n D

Z

240

30 ð 10

3

800

D 10 rev=s or 600 rev=min

Problem 3. An 8-pole, lap-wound armature has 1200 conductors

and a ﬂux per pole of 0.03 Wb. Determine the e.m.f. generated

when running at 500 rev/min.

D.c. machines 355

Generated e.m.f., E D

2pnZ

, for a lap-wound machine, i.e.,

E D nZ D 0.03



500



1200 D 300 volts

Problem 4. Determine the generated e.m.f. in problem 3 if the

armature is wave-wound.

Generated e.m.f. E D

2pnZ

(since c D2 for wave-wound

D pnZ D 4nZ

D 4300 from problem 3,

D 1200 volts

Problem 5. A d.c. shunt-wound generator running at constant

speed generates a voltage of 150 V at a certain value of ﬁeld

current. Determine the change in the generated voltage when the

ﬁeld current is reduced by 20%, assuming the ﬂux is proportional

to the ﬁeld current.

The generated e.m.f. E of a generator is proportional to ω, i.e. is propor-

tional to n, where  is the ﬂux and n is the speed of rotation.

It follows that E D kn, where k is a constant.

At speed n

and ﬂux 

, E

D k

At speed n

and ﬂux 

, E

D k

Thus, by division:

k



The initial conditions are E

D 150 V,  D 

and n D n

. When the

ﬂux is reduced by 20%, the new value of ﬂux is 80/100 or 0.8 of the

initial value, i.e. 

D 0.8

. Since the generator is running at constant

speed, n

D n

Thus



0.8

0.8

that is, E

D 150 ð 0.8 D 120 V

Thus, a reduction of 20% in the value of the ﬂux reduces the generated

voltage to 120 V at constant speed.